Question #33891

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Answer 1 · 2017-01-05T15:53:44

$sf(102.5color(white)(x)s)$

I will consider the motion in two stages.

1. The Powered Stage

To get the final velocity at the end of the powered stage we can use:

$sf(v=u+at)$

This becomes:

$sf(v=0+((9.81)/(3))xx60=196.2color(white)(x)"m/s")$

To find the height h reached we can use:

$sf(v^2=u^2+2as)$

This becomes:

$sf(196.2^2=0+(2xx9.81)/(3)xxh)$

$:.$ $sf(h=(3.849xx10^4xx3)/(2xx9.81)=588.6color(white)(x)m)$

2. The Unpowered Stage

The rocket is now moving under the influence of gravity.

We can use:

$sf(s=ut+1/2at^2)$

I will use the convention that "up is positive".

The equation becomes:

$sf(-588.6=196.2t-1/2xx9.81xxt^2)$

$:.$ $sf(4.9t^2-196.2t-588.6=0)$

Applying the quadratic formula:

$sf(t=(196.2+-sqrt(37,094.76-[4xx4.9xx(-588.6)]))/(9.8)$

$sf(t=(196.2+-220.5)/(9.8))$

Ignoring the -ve root:

$sf(t=42.5color(white)(x)s)$

To get the total time of flight $(sf(t_("tot")))$ , we add this to the time taken for the powered part of the flight which we know is 1 minute.

$:.$ $sf(t_("tot")=60+42.5=102.5color(white)(x)s)$